Concentric circles are circles with a common center. The region between two concentric circles of different radii is called an annulus.
Any two circles can be made concentric by inversion
by picking the inversion center as one of the
limiting points.

Given two concentric circles with radii and , what is the probability that a chord chosen at random from
the outer circle will cut across the inner circle? Depending on how the "random"
chord is chosen, 1/2, 1/3, or 1/4 could all be correct
answers.

1. Picking any two points on the outer circle and connecting them gives 1/3.

2. Picking any random point on a diagonal and then picking the chord
that perpendicularly bisects it gives 1/2.

3. Picking any point on the large circle, drawing a line to the center, and then
drawing the perpendicularly bisected chord gives 1/4.

So some care is obviously needed in specifying what is meant by "random" in this problem.

Given an arbitrary chord to the larger of two concentric circles
centered on ,
the distance between inner and outer intersections is equal on both sides . To prove this, take the perpendicular
to
passing through
and crossing at .
By symmetry, it must be true that and are equal. Similarly, and must be equal. Therefore, equals . Incidentally, this is also true for homeoids,
but the proof is nontrivial.